0580EasyMathStringGreedy

Largest Odd Segment of a Serial

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Trains the technique from

LeetCode 1903Largest Odd Number in String

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A part carries a serial num, written as a run of decimal digits.

A segment of the serial is any unbroken stretch of one or more of those digits, read in the order they are printed. Each segment stands for a whole number; a segment that begins with a zero simply stands for a smaller number.

Among the segments whose value is odd, find the one of greatest value and return it as a string, digit for digit as it is printed on the part. When no segment of the serial is odd, return the empty string.

Examples

Example 1

Input: num = "9146"
Output: "91"

The segment "91" stands for the odd number 91. The only segments of this serial standing for a larger number are "914", "9146" and "146", and none of those three is odd.

Example 2

Input: num = "2468"
Output: ""

Every digit printed here is even, so every segment ends in an even digit and no segment is odd.

Example 3

Input: num = "8003"
Output: "8003"

The whole serial is itself a segment, it stands for 8003, and 8003 is odd. No segment of this serial stands for a larger number.

Constraints

1 <= num.length <= 10^5
num holds decimal digits only.
num has no leading zero unless it is the single digit 0.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def largest_odd_number(num: str) -> str:

Java

public String largestOddNumber(String num)

The intended approach

Time O(n) — a single right-to-left scan plus one slice.Space O(n) for the returned slice, O(1) beyond it.

A segment is odd exactly when its final digit is odd, and among segments the ones that start at the front of the serial are the longest available, so scan inward from the right end for the last odd digit and return the stretch from index 0 through it. If the scan runs off the left end, no segment is odd and the answer is the empty string.

Pythonentry point: largest_odd_number

def largest_odd_number(num):
    # O(n) time, O(n) space for the returned slice: step in from the right end
    # until an odd digit shows up, then hand back everything from the start of
    # the reading up to and including that digit.
    i = len(num) - 1
    while i >= 0 and (ord(num[i]) - 48) % 2 == 0:
        i -= 1
    if i < 0:
        return ""
    return num[:i + 1]

The obvious version

Enumerate every start and every end, convert each segment to an integer, and keep the largest odd value seen along with the digits that produced it.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_largest_odd_number(num):
    best = ""
    best_val = -1
    n = len(num)
    for i in range(n):
        for j in range(i + 1, n + 1):
            block = num[i:j]
            v = int(block)
            if v % 2 == 1 and v > best_val:
                best_val = v
                best = block
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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